HA4 - More demanding factorisation

HA4 - More demanding factorisation

How can you factorise a quadratic when its first term is 6x26x^2 rather than x2x^2? Extend the HA4 algebra methods to nonmonic quadratics, products of three or more brackets and surd coefficients. Use expansion to check that every rewritten expression is identical to the original.

Nonmonic quadratics

For x2+5x+6x^2+5x+6, the pair 2,32,3 has sum five and product six. With 6x2+11x+36x^2+11x+3, looking only for a pair with product three misses the leading coefficient.

Nonmonic quadratic

A nonmonic quadratic has the form ax2+bx+cax^2+bx+c, where a≠0a\ne0 and a≠1a\ne1. To factorise one with integer coefficients, first remove any common factor. Then look for two numbers with product acac and sum bb, split the middle term and factor by grouping.

The split works because (px+q)(rx+s)=prx2+(ps+qr)x+qs(px+q)(rx+s)=prx^2+(ps+qr)x+qs. The two contributions to the middle coefficient, psps and qrqr, have product prqs=acprqs=ac. Not every quadratic has integer or rational linear factors; a failed integer-pair search does not justify inventing a pair.

Worked example 1: a positive constant

Question. Factorise 6x2+11x+36x^2+11x+3.

Plan. Find a pair with product 6(3)=186(3)=18 and sum 1111, then group.

The pair is 9,29,2.

6x2+11x+3=6x2+9x+2x+3=3x(2x+3)+1(2x+3)=(3x+1)(2x+3)\begin{aligned} 6x^2+11x+3&=6x^2+9x+2x+3\\ &=3x(2x+3)+1(2x+3)\\ &=(3x+1)(2x+3) \end{aligned}

Answer. (3x+1)(2x+3)(3x+1)(2x+3).

Check. Expansion gives 6x2+9x+2x+3=6x2+11x+36x^2+9x+2x+3=6x^2+11x+3, including the correct middle term.

Change the constant's sign. The split pair must now have opposite signs.

Worked example 2: a negative constant

Question. Factorise 6x2−7x−36x^2-7x-3.

Plan. Use product −18-18, sum −7-7, then factor each group.

The pair is −9,2-9,2.

6x2−7x−3=6x2−9x+2x−3=3x(2x−3)+1(2x−3)=(3x+1)(2x−3)\begin{aligned} 6x^2-7x-3&=6x^2-9x+2x-3\\ &=3x(2x-3)+1(2x-3)\\ &=(3x+1)(2x-3) \end{aligned}

Answer. (3x+1)(2x−3)(3x+1)(2x-3).

Check. The cross terms are −9x+2x=−7x-9x+2x=-7x; the constant is −3-3.

If the original expression is −12x2+14x+6-12x^2+14x+6, remove −2-2 first to leave −2(6x2−7x−3)-2(6x^2-7x-3). The second example then gives −2(3x+1)(2x−3)-2(3x+1)(2x-3). Retain the removed factor in the final answer.

Watch out

A pair with product acac is only half the requirement: it must also add to bb. For the second example, −6,3-6,3 has product −18-18 but sum −3-3, so it is the wrong pair.

Try a new middle coefficient and constant.

Three brackets

Multiplying three brackets uses the same distributive law as two brackets. Expand two first, collect like terms, then multiply the resulting polynomial by the remaining bracket.

Each term in one factor must multiply every term in the next. Before starting, predict the highest power: three linear factors with leading term xx give x3x^3; four give x4x^4. You can change the order and grouping of multiplication to make the work easier.

Worked example 3: three linear factors

Question. Expand and simplify (x+2)(x−1)(x+3)(x+2)(x-1)(x+3).

Plan. Expand the first pair, then distribute x+3x+3 across all three resulting terms.

(x+2)(x−1)=x2+x−2(x2+x−2)(x+3)=x3+3x2+x2+3x−2x−6=x3+4x2+x−6\begin{aligned} (x+2)(x-1)&=x^2+x-2\\ (x^2+x-2)(x+3)&=x^3+3x^2+x^2+3x-2x-6\\ &=x^3+4x^2+x-6 \end{aligned}

Answer. x3+4x2+x−6x^3+4x^2+x-6.

Check. At x=0x=0, the product is 2(−1)(3)=−62(-1)(3)=-6, matching the constant. At x=1x=1, both forms give zero.

Add a fourth bracket, but pair opposite signs to reduce the amount of expansion.

Continue this lesson.

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